CoffeeCrow
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Homework Statement
1. Compute the electric field vector at the position of charge B for the case where q_{A}=4q_{C} and r_{1}=2r_{2}
2. Suppose that all three charge are positive and that q_{A}=4q_{C}. Consider the case where f q_{B} is moved to the right by a distance dr.
Now r_{1} = 2r+dr and r_{2}=r-dr. Find the electric field vector in terms of q_{C},r,dr and \epsilon_{0} and deduce the direction in which q_{b} will move.
Homework Equations
\overrightarrow{E} = \frac{1}{4\pi \epsilon_{0}}\frac{q}{r^{2}}\hat{r}
The Attempt at a Solution
Part 1 is simple, put everything in terms of r_{2} and q_{c} and add the field due to each charge as vectors\overrightarrow{E}_{b} = \frac {1}{4\pi \epsilon_{0}}(\frac {4q_{c}}{4r^{2}_{2}} - \frac {q_{c}} {r^{2}_{2}}) \hat{i}
\overrightarrow{E}_{b} = 0 \hat{i}
Part 2 is where I'm a little stuck. I try to find the field due to charge A and C as before:
\overrightarrow{E}_{b} = \frac{q_{c}}{4\pi \epsilon_{0}}(\frac{4}{(2r+dr)^{2}}-\frac {1} {(r-dr)^{2}}) \hat{i}
The given answer is:
E=\frac{q_{c}}{4\pi\epsilon_{0}} \frac{-3 dr}{r^{2}(r-dr)}
But no matter how much I manipulate my expression, I can't seem to get anything very close to it. It just gets messier and messier.
I guess I'd just like to ask if my initial expression is correct and I'm just messing up the algebra, or if I've missed something conceptual, or if I have yet a different problem.
Thanks